Sorry for my question but I did not understand the answers that was related to this question so I hope someone can enlighten me further.
I am a new data science student and we are going to learn how to program in the functional language F#. We are learning about algorithms and I wanted to write the algorithms as F# functions to check if my calculations on paper were correct.
I get the following error saying:
"This value is not mutable. Consider using the mutable keyword let mutable n = expression"
My code looks like this:
let loop5( n ) =
let mutable x = 0
while n > 0 do
x <- x + 1
n <- n + 1
printfn "loop5(): x=%i for n=%i" x n
loop5(4)
I'm trying to write a function looking like this (pseudocode):
loop5(n)
x = 0
while n > 0
x = x + 1
n = n + 1
return x
Hope I made a clear question and someone can help me out here :-) Have a nice weekend
You're trying to mutate the loop's parameter n. The parameter is not mutable, so the compiler doesn't let you. That's exactly what the error tells you.
Now, normally, to make the error go away, you'd make the variable mutable. However, you can't make a function parameter mutable, so that's not an option.
Here you want to think what the meaning of your program should be. Does the loop function need to pass the updated value of n back to its caller, or is the whole mutation its internal business? If it's the former, please see #AnyMoose's answer, but from your example and explanation, I suspect it's the latter. If that is the case, simply make a mutable copy of the parameter and work with it:
let loop n' =
let mutable x = 0
let mutable n = n'
...
Separately, I want to point out that your program, as written, would actually loop indefinitely (or until it wraps around the max int value anyway), because instead of decreasing n at each step you're increasing it. If you want your program to actually finish before the next Ice Age, you need to make n decrease with each iteration:
n <- n - 1
Ref cells
Ref cells get around some of the limitations of mutables. In fact, ref cells are very simple datatypes which wrap up a mutable field in a record type. Ref cells are defined by F# as follows:
type 'a ref = { mutable contents : 'a }
The F# library contains several built-in functions and operators for working with ref cells:
let ref v = { contents = v } (* val ref : 'a -> 'a ref *)
let (!) r = r.contents (* val (!) : 'a ref -> 'a *)
let (:=) r v = r.contents <- v (* val (:=) : 'a ref -> 'a -> unit *)
The ref function is used to create a ref cell, the ! operator is used to read the contents of a ref cell, and the := operator is used to assign a ref cell a new value. Here is a sample in fsi:
let x = ref "hello";;
val x : string ref
x;; (* returns ref instance *)
val it : string ref = {contents = "hello";}
!x;; (* returns x.contents *)
val it : string = "hello"
x := "world";; (* updates x.contents with a new value *)
val it : unit = ()
!x;; (* returns x.contents *)
val it : string = "world"
Since ref cells are allocated on the heap, they can be shared across multiple functions:
open System
let withSideEffects x =
x := "assigned from withSideEffects function"
let refTest() =
let msg = ref "hello"
printfn "%s" !msg
let setMsg() =
msg := "world"
setMsg()
printfn "%s" !msg
withSideEffects msg
printfn "%s" !msg
let main() =
refTest()
Console.ReadKey(true) |> ignore
main()
The withSideEffects function has the type val withSideEffects : string ref -> unit.
This program outputs the following:
hello
world
Assigned from withSideEffects function
The withSideEffects function is named as such because it has a side-effect, meaning it can change the state of a variable in other functions. Ref Cells should be treated like fire. Use it cautiously when it is absolutely necessary but avoid it in general. If you find yourself using Ref Cells while translating code from C/C++, then ignore efficiency for a while and see if you can get away without Ref Cells or at worst using mutable. You would often stumble upon a more elegant and more maintanable algorithm
Aliasing Ref Cells
Note: While imperative programming uses aliasing extensively, this practice has a number of problems. In particular it makes programs hard to follow since the state of any variable can be modified at any point elsewhere in an application. Additionally, multithreaded applications sharing mutable state are difficult to reason about since one thread can potentially change the state of a variable in another thread, which can result in a number of subtle errors related to race conditions and dead locks.
A ref cell is very similar to a C or C++ pointer. Its possible to point to two or more ref cells to the same memory address; changes at that memory address will change the state of all ref cells pointing to it. Conceptually, this process looks like this:
Let's say we have 3 ref cells looking at the same address in memory:
Three references to an integer with value 7
cell1, cell2, and cell3 are all pointing to the same address in memory. The .contents property of each cell is 7. Let's say, at some point in our program, we execute the code cell1 := 10, this changes the value in memory to the following:
Three references to an integer with value 10
By assigning cell1.contents a new value, the variables cell2 and cell3 were changed as well. This can be demonstrated using fsi as follows:
let cell1 = ref 7;;
val cell1 : int ref
let cell2 = cell1;;
val cell2 : int ref
let cell3 = cell2;;
val cell3 : int ref
!cell1;;
val it : int = 7
!cell2;;
val it : int = 7
!cell3;;
val it : int = 7
cell1 := 10;;
val it : unit = ()
!cell1;;
val it : int = 10
!cell2;;
val it : int = 10
!cell3;;
val it : int = 10
I can't understand what is wrong with following bit of code:
let toClass (problem:Problem<'a>) (classID:int) (items:'a list) =
let newFreqTable = (problem.FreqTables.[classID]).count items
{ problem with FreqTables = newFreqTable :: (problem.FreqTables |> List.filter (fun i -> i.ClassID <> classID)) }
type Problem<'a> when 'a : equality with member this.toClass (classID:int) (items:list<'a>) = toClass this classID items
I have a Problem type which is nothing but a way to group up any number of FreqTables - short for "Frequency tables". So toClass method just takes appropriate freqTable (by classID argument) and returns a new one - with calculated given items.
let typeIndependentCall = toClass p 0 ["word"; "word"; "s"] // this works perfectly
let typeDependentCall = typeIndependentCall.toClass 1 ["word"; "s"]
// gives an error: "One or more of the overloads of this method has
// curried arguments. Consider redesigning these members to take
// arguments in tupled form".
I am pretty new to F# and functional programming. What is the right way to attach behavior to my type?
In F# there are 2 main ways of passing arguments to a function: curried and tupled. The curried form is what you are using in your code above, and has a few key benefits, the first and foremost being partial application.
For example, instead of thinking of
fun add a b = a + b
as a function that takes in 2 arguments and returns a value, we can think of it as a function of one argument that returns a function that with one argument. This is why the type signature of our function is
Int -> Int -> Int
or, more clearly,
Int -> (Int -> Int)
However, when overloading methods, we can only use the tupled argument form
(Int, Int) -> Int
The reason for this is for optimization, as is discussed here
To get your code to work, use
type Problem<'a> when 'a : equality with member this.toClass (classID:int, items:list<'a>) = toClass this classID items
and call it like such:
let typeDependentCall = typeIndependentCall.toClass(1, ["word"; "s"])
Does Rascal support function pointers or something like this to do this like Java Interfaces?
Essentially I want to extract specific (changing) logic from a common logic block as separate functions. The to be used function is passed to the common block, which then call this function. In C we can do this with function pointers or with Interfaces in Java.
First I want to know how this general concept is called in the language design world.
I checked the Rascal Function Helppage, but this provide no clarification on this aspect.
So e.g. I have:
int getValue(str input) {
.... }
int getValue2(str input){
... }
Now I want to say:
WhatDatatype? func = getValue2; // how to do this?
Now I can pass this to an another function and then:
int val = invoke_function(func,"Hello"); // how to invoke?, and pass parameters and get ret value
Tx,
Jos
This page in the tutor has an example of using higher-order functions, which are the Rascal feature closest to function pointers:
http://tutor.rascal-mpl.org/Rascal/Rascal.html#/Rascal/Concepts/Functions/Functions.html
You can define anonymous (unnamed) functions, called closures in Java; assign them to variables; pass them as arguments to functions (higher-order functions); etc. Here is an example:
rascal>myfun = int(int x) { return x + 1; };
int (int): int (int);
rascal>myfun;
int (int): int (int);
rascal>myfun(3);
int: 4
rascal>int applyIntFun(int(int) f, int x) { return f(x); }
int (int (int), int): int applyIntFun(int (int), int);
rascal>applyIntFun(myfun,10);
int: 11
The first command defines an increment function, int(int x) { return x + 1; }, and assigns this to variable myfun. The rest of the code would work the same if instead this was
int myfun(int x) { return x + 1; }
The second command just shows the type, which is a function that takes and returns int. The third command calls the function with value 3, returning 4. The fourth command then shows a function which takes a function as a parameter. This function parameter, f, will then be called with argument x. The final command just shows an example of using it.
let GetVal (i,isMin,al, be)=
let b = new Board(board)
if b.SetBoardBool(i) then this.MinMaxAlphaBeta(b, isMin, al, be)
else -2
let valList = seq{
for i =0 to 8 do
yield (GetVal i (not isMin) alphaF betaF , not isMin)
}
I am getting an F# error saying: This value is not a function and cannot be applied.
valList is sequence of tuples of int and bool and GetVal takes int bool int int and returns int. where alphaF betaF are mutable variables.
Or you could change the signature of GetVal to not pass a tuple--like this:
let GetVal i isMin al be =
i, isMin, al, and be are called curried parameters. You can find more detail here under the topic "Partial Application of Arguments." I would post a direct link but there doesn't seem to be one.
Your GetVal function takes tupled arguments (a,b,c,d) whilst you call it with curried arguments a b c d
Something like this should work
yield (GetVal (i, (not isMin), alphaF, betaF) , not isMin)
I wonder what this means in F#.
“a function taking an integer, which returns a function which takes an integer and returns an integer.”
But I don't understand this well.
Can anyone explain this so clear ?
[Update]:
> let f1 x y = x+y ;;
val f1 : int -> int -> int
What this mean ?
F# types
Let's begin from the beginning.
F# uses the colon (:) notation to indicate types of things. Let's say you define a value of type int:
let myNumber = 5
F# Interactive will understand that myNumber is an integer, and will tell you this by:
myNumber : int
which is read as
myNumber is of type int
F# functional types
So far so good. Let's introduce something else, functional types. A functional type is simply the type of a function. F# uses -> to denote a functional type. This arrow symbolizes that what is written on its left-hand side is transformed into what is written into its right-hand side.
Let's consider a simple function, that takes one argument and transforms it into one output. An example of such a function would be:
isEven : int -> bool
This introduces the name of the function (on the left of the :), and its type. This line can be read in English as:
isEven is of type function that transforms an int into a bool.
Note that to correctly interpret what is being said, you should make a short pause just after the part "is of type", and then read the rest of the sentence at once, without pausing.
In F# functions are values
In F#, functions are (almost) no more special than ordinary types. They are things that you can pass around to functions, return from functions, just like bools, ints or strings.
So if you have:
myNumber : int
isEven : int -> bool
You should consider int and int -> bool as two entities of the same kind: types. Here, myNumber is a value of type int, and isEven is a value of type int -> bool (this is what I'm trying to symbolize when I talk about the short pause above).
Function application
Values of types that contain -> happens to be also called functions, and have special powers: you can apply a function to a value. So, for example,
isEven myNumber
means that you are applying the function called isEven to the value myNumber. As you can expect by inspecting the type of isEven, it will return a boolean value. If you have correctly implemented isEven, it would obviously return false.
A function that returns a value of a functional type
Let's define a generic function to determine is an integer is multiple of some other integer. We can imagine that our function's type will be (the parenthesis are here to help you understand, they might or might not be present, they have a special meaning):
isMultipleOf : int -> (int -> bool)
As you can guess, this is read as:
isMultipleOf is of type (PAUSE) function that transforms an int into (PAUSE) function that transforms an int into a bool.
(here the (PAUSE) denote the pauses when reading out loud).
We will define this function later. Before that, let's see how we can use it:
let isEven = isMultipleOf 2
F# interactive would answer:
isEven : int -> bool
which is read as
isEven is of type int -> bool
Here, isEven has type int -> bool, since we have just given the value 2 (int) to isMultipleOf, which, as we have already seen, transforms an int into an int -> bool.
We can view this function isMultipleOf as a sort of function creator.
Definition of isMultipleOf
So now let's define this mystical function-creating function.
let isMultipleOf n x =
(x % n) = 0
Easy, huh?
If you type this into F# Interactive, it will answer:
isMultipleOf : int -> int -> bool
Where are the parenthesis?
Note that there are no parenthesis. This is not particularly important for you now. Just remember that the arrows are right associative. That is, if you have
a -> b -> c
you should interpret it as
a -> (b -> c)
The right in right associative means that you should interpret as if there were parenthesis around the rightmost operator. So:
a -> b -> c -> d
should be interpreted as
a -> (b -> (c -> d))
Usages of isMultipleOf
So, as you have seen, we can use isMultipleOf to create new functions:
let isEven = isMultipleOf 2
let isOdd = not << isEven
let isMultipleOfThree = isMultipleOf 3
let endsWithZero = isMultipleOf 10
F# Interactive would respond:
isEven : int -> bool
isOdd : int -> bool
isMultipleOfThree : int -> bool
endsWithZero : int -> bool
But you can use it differently. If you don't want to (or need to) create a new function, you can use it as follows:
isMultipleOf 10 150
This would return true, as 150 is multiple of 10. This is exactly the same as create the function endsWithZero and then applying it to the value 150.
Actually, function application is left associative, which means that the line above should be interpreted as:
(isMultipleOf 10) 150
That is, you put the parenthesis around the leftmost function application.
Now, if you can understand all this, your example (which is the canonical CreateAdder) should be trivial!
Sometime ago someone asked this question which deals with exactly the same concept, but in Javascript. In my answer I give two canonical examples (CreateAdder, CreateMultiplier) inf Javascript, that are somewhat more explicit about returning functions.
I hope this helps.
The canonical example of this is probably an "adder creator" - a function which, given a number (e.g. 3) returns another function which takes an integer and adds the first number to it.
So, for example, in pseudo-code
x = CreateAdder(3)
x(5) // returns 8
x(10) // returns 13
CreateAdder(20)(30) // returns 50
I'm not quite comfortable enough in F# to try to write it without checking it, but the C# would be something like:
public static Func<int, int> CreateAdder(int amountToAdd)
{
return x => x + amountToAdd;
}
Does that help?
EDIT: As Bruno noted, the example you've given in your question is exactly the example I've given C# code for, so the above pseudocode would become:
let x = f1 3
x 5 // Result: 8
x 10 // Result: 13
f1 20 30 // Result: 50
It's a function that takes an integer and returns a function that takes an integer and returns an integer.
This is functionally equivalent to a function that takes two integers and returns an integer. This way of treating functions that take multiple parameters is common in functional languages and makes it easy to partially apply a function on a value.
For example, assume there's an add function that takes two integers and adds them together:
let add x y = x + y
You have a list and you want to add 10 to each item. You'd partially apply add function to the value 10. It would bind one of the parameters to 10 and leaves the other argument unbound.
let list = [1;2;3;4]
let listPlusTen = List.map (add 10)
This trick makes composing functions very easy and makes them very reusable. As you can see, you don't need to write another function that adds 10 to the list items to pass it to map. You have just reused the add function.
You usually interpret this as a function that takes two integers and returns an integer.
You should read about currying.
a function taking an integer, which returns a function which takes an integer and returns an integer
The last part of that:
a function which takes an integer and returns an integer
should be rather simple, C# example:
public int Test(int takesAnInteger) { return 0; }
So we're left with
a function taking an integer, which returns (a function like the one above)
C# again:
public int Test(int takesAnInteger) { return 0; }
public int Test2(int takesAnInteger) { return 1; }
public Func<int,int> Test(int takesAnInteger) {
if(takesAnInteger == 0) {
return Test;
} else {
return Test2;
}
}
You may want to read
F# function types: fun with tuples and currying
In F# (and many other functional languages), there's a concept called curried functions. This is what you're seeing. Essentially, every function takes one argument and returns one value.
This seems a bit confusing at first, because you can write let add x y = x + y and it appears to add two arguments. But actually, the original add function only takes the argument x. When you apply it, it returns a function that takes one argument (y) and has the x value already filled in. When you then apply that function, it returns the desired integer.
This is shown in the type signature. Think of the arrow in a type signature as meaning "takes the thing on my left side and returns the thing on my right side". In the type int -> int -> int, this means that it takes an argument of type int — an integer — and returns a function of type int -> int — a function that takes an integer and returns an integer. You'll notice that this precisely matches the description of how curried functions work above.
Example:
let f b a = pown a b //f a b = a^b
is a function that takes an int (the exponent) and returns a function that raises its argument to that exponent, like
let sqr = f 2
or
let tothepowerofthree = f 3
so
sqr 5 = 25
tothepowerofthree 3 = 27
The concept is called Higher Order Function and quite common to functional programming.
Functions themselves are just another type of data. Hence you can write functions that return other functions. Of course you can still have a function that takes an int as parameter and returns something else. Combine the two and consider the following example (in python):
def mult_by(a):
def _mult_by(x):
return x*a
return mult_by
mult_by_3 = mult_by(3)
print mylt_by_3(3)
9
(sorry for using python, but i don't know f#)
There are already lots of answers here, but I'd like to offer another take. Sometimes explaining the same thing in lots of different ways helps you to 'grok' it.
I like to think of functions as "you give me something, and I'll give you something else back"
So a Func<int, string> says "you give me an int, and I'll give you a string".
I also find it easier to think in terms of 'later' : "When you give me an int, I'll give you a string". This is especially important when you see things like myfunc = x => y => x + y ("When you give curried an x, you get back something which when you give it a y will return x + y").
(By the way, I'm assuming you're familiar with C# here)
So we could express your int -> int -> int example as Func<int, Func<int, int>>.
Another way that I look at int -> int -> int is that you peel away each element from the left by providing an argument of the appropriate type. And when you have no more ->'s, you're out of 'laters' and you get a value.
(Just for fun), you can transform a function which takes all it's arguments in one go into one which takes them 'progressively' (the official term for applying them progressively is 'partial application'), this is called 'currying':
static void Main()
{
//define a simple add function
Func<int, int, int> add = (a, b) => a + b;
//curry so we can apply one parameter at a time
var curried = Curry(add);
//'build' an incrementer out of our add function
var inc = curried(1); // (var inc = Curry(add)(1) works here too)
Console.WriteLine(inc(5)); // returns 6
Console.ReadKey();
}
static Func<T, Func<T, T>> Curry<T>(Func<T, T, T> f)
{
return a => b => f(a, b);
}
Here is my 2 c. By default F# functions enable partial application or currying. This means when you define this:
let adder a b = a + b;;
You are defining a function that takes and integer and returns a function that takes an integer and returns an integer or int -> int -> int. Currying then allows you partiallly apply a function to create another function:
let twoadder = adder 2;;
//val it: int -> int
The above code predifined a to 2, so that whenever you call twoadder 3 it will simply add two to the argument.
The syntax where the function parameters are separated by space is equivalent to this lambda syntax:
let adder = fun a -> fun b -> a + b;;
Which is a more readable way to figure out that the two functions are actually chained.